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Lecture 2: Trigonometric Functions: Identities
2.1 Some basic identities
For any angle θ, (cos(θ), sin(θ)) is a point on the unit circle x2 + y 2 = 1. Hence we must
have
sin2 (θ) + cos2 (θ) = 1
for all angles θ.
Example
If sin(θ) = 14 , then
cos2 (θ) = 1 − sin2 (θ) = 1 −
Thus
1
15
=
.
16
16
√
15
4
or
√
15
cos(θ) = −
.
4
Since sin(θ) > 0, θ must lie in either the first or second quadrant. If in the first, then
√
15
;
cos(θ) =
4
cos(θ) =
if in the second, then
√
cos(θ) = −
15
.
4
Dividing both sides of
sin2 (θ) + cos2 (θ) = 1
by cos2 (θ) gives us
tan2 (θ) + 1 = sec2 (θ)
for all θ.
The geometry of the circle shows us that, for any angle θ, the point (cos(−θ), sin(−θ)) is
the point (cos(θ), sin(θ)) reflected over the x-axis, an action which does not change the
value of the x-coordinate, but does change the sign of the y-coordinate. That is,
cos(−θ) = cos(θ)
and
sin(−θ) = − sin(θ)
for all θ.
2-1
Lecture 2: Trigonometric Functions: Identities
Example
We have
π
π √3
cos −
= cos
=
6
6
2
and
π
π
1
sin −
= − sin
=− .
6
6
2
Since the unit circle has circumference 2π, we have
sin(θ + 2nπ) = sin(θ)
and
cos(θ + 2nπ) = cos(θ)
for any integer n. We say that the functions sin(θ) and cos(θ) have period 2π.
Example
We have
sin
and
cos
8π
3
8π
3
= sin
8π
− 2π
3
= cos
8π
− 2π
3
√
= sin
2π
3
=
= cos
2π
3
1
=− .
2
3
2
2.2 Addition and subtraction formulas
For any real numbers x and y,
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
and
sin(x + y) = sin(x) cos(y) + cos(x) sin(y).
These are the addition formulas for sine and cosine.
Example
Since
π π
3π 4π
7π
+ =
+
=
,
4
3
12
12
12
it follows that
cos
7π
12
= cos
π
4
cos
π
3
− sin
π
4
sin
π
3
√
√
1 3
1− 3
1 1
√
−√
=
=√
2 2
2 2
22
2-2
Lecture 2: Trigonometric Functions: Identities
2-3
and
sin
7π
12
= sin
π
4
cos
π
3
+ cos
π
4
sin
π
3
√
√
1+ 3
1 1
1 3
√ .
=
=√
+√
22
2 2
2 2
It follows from the addition formulas that for any real numbers x and y,
cos(x − y) = cos(x) cos(y) + sin(x) sin(y)
and
sin(x − y) = sin(x) cos(y) − cos(x) sin(y).
These are the subtraction formulas for sine and cosine.
Example
Since
3π 4π
π
π π
− =
−
=− ,
4
3
12
12
12
it follows that
√
√
π
π
π
π
π
1 1
1 3
1+ 3
√
= cos
cos
+ sin
sin
=√
+√
=
cos −
12
4
3
4
3
22
2 2
2 2
and
√
√
π
π
π
π
π
1 1
1 3
1− 3
√ .
sin −
= sin
cos
− cos
sin
=√
−√
=
12
4
3
4
3
22
2 2
2 2
2.3 Double and half angle formulas
From the addition formula for cosine we have, for any real number x,
cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2 (x) − sin2 (x)
and from the addition formula for sine we have, for any real number x,
sin(2x) = sin(x + x) = sin(x) cos(x) + cos(x) sin(x) = 2 sin(x) cos(x),
the double angle formulas for sine and cosine. Now sin2 (x) = 1 − cos2 (x), so from the
double angle formula for cosine we have
cos(2x) = cos2 (x) − (1 − cos2 (x)) = 2 cos2 (x) − 1.
Solving for cos(x), we have
cos2 (x) =
1 + cos(2x)
,
2
Lecture 2: Trigonometric Functions: Identities
2-4
the half-angle formula for cosine. Similarly, using cos2 (x) = 1 − sin2 (x), we have
cos(2x) = (1 − sin2 (x)) − sin2 (x) = 1 − 2 sin2 (x),
from which we obtain
sin2 (x) =
1 − cos(2x)
,
2
the half-angle formula for sine.
Example
We have
π
π 1 + cos
2
6
cos
=
12
2
and so
π
cos
=
12
√
3
√
2
+
3
2 =
,
2
4
1+
=
p
2+
2
√
Similarly,
π
π 1 − cos
6
=
sin2
12
2
3
.
√
3
√
2 = 2 − 3,
2
4
1−
=
and so
p
√
π
2− 3
=
.
sin
12
2
Note that
we used the positive square roots because we know that both cos
π
sin 12 are positive.
π
12
and
2.4 Shift formulas
We may also derive a number of shift formulas from the addition and subtractions formulas.
For example,
π
π
π
cos x −
= cos(x) cos
+ sin(x) sin
= sin(x)
2
2
2
for all x and
π
π
π
sin x −
= sin(x) cos
− cos(x) sin
= − cos(x)
2
2
2
for all x. Also,
cos(x + π) = cos(x) cos(π) − sin(x) sin(π) = − cos(x)
for all x and
sin(x + π) = sin(x) cos(π) + cos(x) sin(π) = − sin(x)
for all x. As a consequence, note that
− sin(x)
sin(x + π)
tan(x + π) =
=
= tan(x)
cos(x + π)
− cos(x)
for all x. That is, tan(x) has period π.
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